y^{''}-4y^'+4y=t^{-6}e^{2t}

Lösung

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{- 6} e^{2 t}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t}}{20 t ^{4}}

Schritte zur Lösung

y^{''} (t) - 4 y^{'} (t) + 4 y = t^{- 6} e^{2 t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t}}{20 t ^{4}}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t}}{20 t ^{4}}

\frac{\partial}{\partial y} (x e^{x^{2} y})

y^{^{''}} - 4 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 5

y^{^{''}} - y = e^{t}

f^{^{'}} (x) = \frac{( 5 x + 1 )}{5 x}

\int_{0}^{1} 2 x - x^{2} d x